Question

An unbiased die, with faces numbered $1,2,3,4,5,6,$ is thrown $n$ times and the list of $n$ numbers showing up is noted. What is the probability that, among the numbers $1,2,3,4,5,6$, only three numbers appear in this list?

• A. ${}^6{C}_3\frac{(3^n-3\cdot 2^n+3)}{6^n}$
• B. ${}^6{C}_3\frac{(3^n+3\cdot 2^n+3)}{6^n}$
• C. ${}^6{C}_3\frac{(3^n-3\cdot 2^n+3)}{3^n}$
• D. None of these

Answer 1

The correct option is A ${}^6{C}_3\frac{(3^n-3\cdot 2^n+3)}{6^n}$

Let us define a onto function $F$ from $A:[r_1,r_2,...r_n]$ to $B:[1,2,3]$ where $r_1,r_2,...r_n$, are the readings of $n$ throws and $1,2,3$ are the numbers that appear in the $n$ throws.

Number of such functions, $M=N-[n\left(1\right)-n\left(z\right)+n\left(3\right)]$ where $N=$ total number of functions and

$n\left(t\right)=numberoffunctionhavingexactly$ $t$ elements in the range.

Now,$N=3^n$

$n\left(1\right)=3.2^n$

$n\left(2\right)=3$

$n\left(3\right)=0$

$\Rightarrow M=3^n-3.2^n+3$

Hence, the total number of favourable cases

$={(3^n-3.2^n+3)}^6{C}_3$

$\therefore$Required probability$=\frac{{(3^n-3.2^n+3)}^6{C}_3}{6^n}$